What is the solution for y^{''}-2y^'+5y=0,y(0)=2,y^'(0)=-2 ?

The solution for y^{''}-2y^'+5y=0,y(0)=2,y^'(0)=-2 is y=e^t(2cos(2t)-2sin(2t))

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y^{''}-2y^'+5y=0,y(0)=2,y^'(0)=-2

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = - 2

y = e^{t} (2 cos (2 t) - 2 sin (2 t))

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{t} (2 cos (2 t) - 2 sin (2 t))

y = e^{t} (2 cos (2 t) - 2 sin (2 t))

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = α, y^{^{'}} (0) = 5

49 x^{^{''}} + 140 x^{^{'}} + 100 x = 0

35 y^{^{''}} + 23 y^{^{'}} - 6 y = 0

y^{^{''}} + 25 y = 10 y

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (π) = 6, y (0) = 6

What is the solution for y^{''}-2y^'+5y=0,y(0)=2,y^'(0)=-2 ?
The solution for y^{''}-2y^'+5y=0,y(0)=2,y^'(0)=-2 is y=e^t(2cos(2t)-2sin(2t))